[水题]ZOJ3038 Triangle War II

  1 struct node
  2 {
  3     bool a[25];
  4 }front, rear;
  5 queue<node> q;
  6 int ans;
  7 bool vis[2100000];
  8 int biao[]={0, 1, 3, 6, 10, 15, 21};
  9 int b[][2]={
 10             {0,0},
 11             {-1,0},
 12             {-1,-1},
 13             {1,0},
 14             {1,1},
 15             {0,1},
 16             {0,-1}
 17            };
 18 int l[25], c[25][10];
 19 void pre()
 20 {
 21     memset(l, 0, sizeof(l));
 22     for(int i=1;i<=21;i++)
 23     {
 24         int x=0;
 25         while(biao[x]<i)
 26             x++;
 27         int y=i-x*(x-1)/2;
 28         for(int k=0;k<7;k++)
 29         {
 30             int dx=x+b[k][0];
 31             int dy=y+b[k][1];
 32             if(dx>=1 && dx<=6 && dy>=1 && dy<=dx)
 33                 c[i][l[i]++]=dx*(dx-1)/2+dy;
 34         }
 35     }
 36 }
 37 void bfs(int n)
 38 {
 39     int d=0;
 40     for(int i=1;i<=n;i++)
 41         d=(d<<1)+front.a[i];
 42     vis[d]=1;
 43     q.push(front);
 44     while(!q.empty())
 45     {
 46         front=q.front();
 47         q.pop();
 48         for(int i=1;i<=n;i++)
 49             if(!front.a[i])
 50             {
 51                 rear=front;
 52                 for(int j=0;j<l[i];j++)
 53                     rear.a[c[i][j]]=1^front.a[c[i][j]];
 54                 d=0;
 55                 int cnt=0;
 56                 for(int j=1;j<=n;j++)
 57                 {
 58                     d=(d<<1)+rear.a[j];
 59                     if(!rear.a[j])
 60                         cnt++;
 61                 }
 62                 if(!vis[d])
 63                 {
 64                     vis[d]=1;
 65                     ans=max(ans, cnt);
 66                     if(ans==n)
 67                         return ;
 68                     q.push(rear);
 69                 }
 70             }
 71     }
 72 }
 73 int main()
 74 {
 75     pre();
 76     int n;
 77     while(~scanf("%d", &n))
 78     {
 79         ans=0;
 80         for(int i=1;i<=n;i++)
 81             for(int j=1;j<=i;j++)
 82             {
 83                 char ch;
 84                 cin>>ch;
 85                 if(ch==‘.‘)
 86                     ans++;
 87                 front.a[i*(i-1)/2+j]=(ch==‘*‘);
 88             }
 89         n=(n+1)*n/2;
 90         if(ans==n)
 91         {
 92             printf("%d\n", ans);
 93             continue;
 94         }
 95         while(!q.empty())
 96             q.pop();
 97         memset(vis, 0, sizeof(vis));
 98         bfs(n);
 99         printf("%d\n", ans);
100     }
101     return 0;
102 }